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Homework Help: E=KE+PE, x,v,a calc question

  1. Apr 22, 2013 #1


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    This problem is from the 2009 AP C Mechanics Exam. I get the same answers as the posted solutions except for part d. I'm hoping someone can tell me what I'm doing wrong. I'm going to show my work for part a and part c even though I got them correct, because they are used for part d.

    A 3.0 kg object is moving along the x-axis in a region where its potential energy as a function of x is given as , where U is in joules and x is in meters. When the object passes the point x = -0.50 m, its velocity is + 2.0 m/s. All forces acting on the object are conservative.

    (a) Calculate the total mechanical energy of the object.
    E = KE + PE = \frac{1}{2}mv^2 + 4x^2 \\
    = \frac{1}{2}\left( {3\,{\rm{kg}}} \right)\left( {2\,{\rm{m/s}}} \right)^2 + 4\left( { - 0.5\,{\rm{m}}} \right)^2 =7 J \\
    (b) Calculate the x-coordinate of any points at which the object has zero kinetic energy.
    (c) Calculate the magnitude of the momentum of the object at x = 0.60 m.
    Come up with a formula for velocity as a function of time, using the formula derived in part a. This velocity formula will also be used in part d.
    KE = \frac{1}{2}mv^2 = E - U = 7 - 4x^2 \\
    v = \sqrt {\frac{{14 - 8x^2 }}{m}} = \sqrt {\frac{{14 - 8\left( {0.6} \right)^2 }}{3}} = 1.93\,{\rm{m/s}} \\
    p = mv = \left( 3 \right)\left( {1.93} \right) = 5.77\,{\rm{kg - m/s}} \\

    (d) Calculate the magnitude of the acceleration of the object as it passes x = 0.60 m.
    acceleration is the derivative of velocity, so take the derivative and set t to 0.6.
    KE\left( x \right) = 7 - U\left( x \right) = 7 - 4x^2 \\
    v\left( x \right) = \sqrt {\frac{{2\left( {7 - 4x^2 } \right)}}{m}} \\
    a\left( x \right) = \frac{1}{2}\frac{2}{m}\left( { - 8x} \right)\left[ {\frac{{2\left( {7 - 4x^2 } \right)}}{m}} \right]^{ - 1/2} \\
    \frac{{ - 8x}}{m}\left[ {\frac{{2\left( {7 - 4x^2 } \right)}}{m}} \right]^{ - 1/2} \\
    \frac{{ - 8\left( {0.6} \right)}}{3}\left[ {\frac{{2\left( {7 - 4\left( {0.6} \right)^2 } \right)}}{3}} \right]^{ - 1/2} =-0.83 m/s^{2} \\

    But this is not the same answer as the scoring guide. They get 1.6 m/s2.
    I know my velocity formula is correct, because I got the correct momentum in part c. I'm pretty sure I took the derivative correctly. I even verified that with graphing software.

    Additionally, I did a double check by using my velocity formula to compute the velocity at 0.6 s and 0.6001 s. The acceleration would then be (v2-v1)/Δx , where Δx is 0.0001. I also get -0.83 m/s2 with this method.

    How did I get the same wrong answer using 2 different methods?

    Their justification is here: http://apcentral.collegeboard.com/apc/public/repository/ap09_physics_c_mechanics_sgs.pdf

  2. jcsd
  3. Apr 22, 2013 #2


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    Yeah, tricky. I think you're the one who's in the wrong though, unfortunately. You solved part d by assuming this:

    ##a(x) = \frac{d}{dx}v(x)## × WRONG

    What's actually true is that acceleration is the TIME derivative of velocity:$$a(x) = \frac{d}{dt}v(x)$$ and so you have to differentiate using the chain rule, just like they did in the solutions manual.

    Your first method was to calculate dv/dx exactly, and your second method was to calculate Δv/Δx for a small but finite Δx of 0.0001. Both of these were wrong, because neither of them is the acceleration, which is dv/dt (instantaneous) or Δv/Δt (average).

  4. Apr 22, 2013 #3

    Doc Al

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    Redo your derivative. Looks like you messed up when applying the chain rule. And as cepheid explains, you need the derivative with respect to time to do it your way.

    The easy way to find the acceleration is to first find the force: F = - dU/dx
  5. Apr 22, 2013 #4


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    I was pretty sure I was the one who got it wrong. If they made a mistake, they would have noticed it after grading a few thousand exams! Thanks Cephid and Doc Al. I could have stared at that for another hour and not noticed my wrong variable!
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